Question: Philip ran out of time while taking a multiple-choice test and plans to guess on the last $4$ questions. Each question has $5$ possible choices, one of which is correct. Let $X=$ the number of answers Philip correctly guesses in the last $4$ questions. Assume that the results of his guesses are independent. What is the probability that he answers exactly $1$ question correctly in the last $4$ questions? You may round your answer to the nearest hundredth. $P(X=1)=$
Without a fancy calculator Answering $1$ question correctly in the last $4$ questions means Philip needs to get $1$ question correct and $3$ questions not correct. There are $5$ possible choices for each question, so we know $P({\text{correct}})={20\%}$ and $P({\text{not}})={80\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of getting $1$ question correct followed by $3$ questions not correct: $P({\text{C}}{\text{NNN}})=({0.2})({0.8})^3=0.1024$ This isn't our final answer, because there are other ways to get $1$ question correct in $4$ questions (for example, NNNC). How many different ways are there? We can use the combination formula to find how many ways there are to get $1$ question correct in $4$ questions: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _4\text{C}_1&=\dfrac{4!}{(4-1)!\cdot1!} \\\\ &=\dfrac{4 \cdot \cancel{3 \cdot 2 \cdot 1}}{(\cancel{3 \cdot 2 \cdot 1}) \cdot 1} \\\\ &=4 \end{aligned}$ There are $4$ ways to get $1$ question correct in $4$ questions. Do they all have the same probability? Each of the $4$ ways has the same probability that we already found: $\begin{aligned} P({\text{C}}{\text{NNN}})&=({0.2})({0.8})^3=0.1024 \\\\ P({\text{N}}{\text{C}}{\text{NN}})&=({0.2})({0.8})^3=0.1024 \\\\ P({\text{NN}}{\text{C}}{\text{N}})&=({0.2})({0.8})^3=0.1024 \\\\ P({\text{NNN}}{\text{C}})&=({0.2})({0.8})^3=0.1024 \end{aligned}$ So we can multiply this probability by $4$ since that is how many ways there are to get $1$ question correct in $4$ questions: $\begin{aligned} P(X=1)&=4(0.2)(0.8)^3 \\\\ &=4(0.1024) \\\\ &=0.4096 \\\\ &\approx0.41 \end{aligned}$ Answer $P(X=1)=0.4096\approx0.41$